Ch20-p031 - 31(a The net work done is the rectangular area enclosed in the pV diagram W =(V V0 p p0 = 2V0 V0 2 p0 p0 = V0 p0 Inserting the values

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1 1 1 0.750 75.0% 4 a c T T ε =− =− = = where the gas law in ratio form has been used. (e) This is greater than our result in part (c), as expected from the second law of thermodynamics. 31. (a) The net work done is the rectangular “area” enclosed in the pV diagram: () ( ) ( ) 00 0 0 0 0 0 0 22 . WV Vp p V V pp V p =− = = Inserting the values stated in the problem, we obtain W = 2.27 kJ. (b) We compute the energy added as heat during the “heat-intake” portions of the cycle using Eq. 19-39, Eq. 19-43, and Eq. 19-46: ()() 35 +1 + 3 5 1+ 2 1+ 4 2 2 2 13 2 bc b abc V b a p c b a a aa a b a a TT T Qn C T T n C T T n R T n R T T T nRT p V T pV §· § · = ¨¸ ¨ ¸
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This note was uploaded on 06/03/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.

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